Discrete Mathematics

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1 Dscrete Mathematcs 3 (0) 6 40 Contents lsts avalable at ScVerse ScenceDrect Dscrete Mathematcs journal homepage: Hamltonan cycles wth all small even chords Guantao Chen a, Katsuhro Ota b, Akra Sato c, Y Zhao a, a Department of Mathematcs and Statstcs, Georga State Unversty, Atlanta, GA 30303, Unted States b Department of Mathematcs, Keo Unversty, 3-4- Hyosh, Kohoku-Ku, Yokohama 3-85, Japan c Department of Computer Scence, Nhon Unversty, Sakurajosu , Setagaya-Ku, Tokyo , Japan a r t c l e n f o a b s t r a c t Artcle hstory: Receved 7 December 00 Receved n revsed form 8 December 0 Accepted 9 December 0 Keywords: Hamltonan cycle Drac theorem Posa s conjecture Let G be a graph of order n 3. An even squared Hamltonan cycle (ESHC) of G s a Hamltonan cycle C = v v... v n v of G wth chords v v +3 for all n (where v n+j = v j for j ). When n s even, an ESHC contans all bpartte -regular graphs of order n. We prove that there s a postve nteger N such that for every graph G of even order n N, f the mnmum degree s δ(g) n + 9, then G contans an ESHC. We show that the condton of n beng even cannot be dropped and the constant 9 cannot be replaced by. Our results can be easly extended to even kth powered Hamltonan cycles for all k. 0 Elsever B.V. All rghts reserved.. Introducton In ths paper, we wll only consder smple graphs fnte graphs wthout loops or multple edges. The notatons and defntons not defned here can be found n [7]. Let G = (V, E) be a graph wth vertex set V and edge set E. For a vertex v V and a subset S V, let Γ (v, S) denote the set of neghbors of v n S, and deg(v, S) = Γ (v, S). Gven another set U V, defne Γ (U, S) = u U Γ (u, S) and deg(u, S) = Γ (U, S). When U = {v,..., v k }, we smply wrte Γ (U, S) and deg(u, S) as Γ (v,..., v k, S) and deg(v,..., v k, S), respectvely. When S = V, we only wrte Γ (U) and deg(u). A graph G s called Hamltonan f t contans a spannng cycle. The Hamltonan problem, determnng whether a graph has a Hamltonan cycle, has long been one of few fundamental problems n graph theory. In ths paper, we fx G to be a graph of order n 3. Drac [8] proved that, f the mnmum degree δ(g) n/ then G s Hamltonan. Ore [3] extended Drac s result by replacng the mnmum degree condton wth that of deg(u) + deg(v) n for all nonadjacent vertces u and v. Many results have been obtaned on generalzng these two classc results (see [3] for a recent survey n ths area). A -regular subgraph (-factor) of G conssts of dsjont cycles of G. Agner and Brandt [] proved that f the mnmum degree δ(g) n then G contans all -factors as subgraphs (Alon and Fscher [3] proved ths for suffcently large n). 3 If to-be-embedded -factors have at most k odd components, then by a conjecture of El-Zahar [9], the mnmum degree condton can be reduced to δ(g) (n + k)/ (Abbas [] announced a proof of El-Zahar s conjecture for large n). Another way to generalze Agner and Brandt s result s to fnd one specfc subgraph of G that contans all -factors of G. A squared Hamltonan cycle of G s a Hamltonan cycle v v v n v together wth edges v v + for all n. Note that we always assume that v n+ = v for. It s easy to see that a squared Hamltonan cycle contans all -factors of G. Pósa (see [0]) conjectured that every graph G of order n 3 wth δ(g) n contans a squared Hamltonan cycle. Fan and Kerstead [] 3 proved ths conjecture approxmately; Komlós et al. [5] proved the conjecture for suffcently large n. More generally, the k-th powered Hamltonan cycle s a Hamltonan cycle v v v n v wth chords v v +j for all n and j k. Correspondng author. E-mal addresses: gchen@gsu.edu (G. Chen), yzhao6@gsu.edu, matyxz@langate.gsu.edu (Y. Zhao) X/$ see front matter 0 Elsever B.V. All rghts reserved. do:0.06/j.dsc.0..03

2 G. Chen et al. / Dscrete Mathematcs 3 (0) Komlós et al. [7,8] proved a conjecture of Seymour for suffcently large n: every n-vertex graph G wth δ(g) (k )n/k contans a kth powered Hamltonan cycle. Böttcher et al. [4] recently proved a conjecture of Bollobás and Komlós (see [4]), whch asymptotcally ncludes all the results mentoned above. Gven an nteger b, a graph H s sad to have bandwdth at most b, f there exsts a labelng of the vertces by v, v,..., v n, such that j b whenever v v j E(H). It s shown n [4] that for any ε > 0 and ntegers r,, there exsts β > 0 wth the followng property. Let G and H be n-vertex graphs for suffcently large n. If δ(g) ((r )/r + ε)n and H s r-chromatc wth maxmum degree and bandwdth at most βn, then G contans a copy of H. Note that the kth powered Hamltonan cycle of order n has chromatc number k+ or k+ dependng on the value of n. The authors of [4] make ther result applcable even when H s r + -chromatc but one of ts color classes s farly small, e.g., the kth powered Hamltonan cycle. We are nterested n the stuaton when the error term εn n the conjecture of Bollobás and Komlós can be reduced to a constant. Accordng to the El-Zahar Conjecture, every n-vertex graph G wth the mnmum degree δ(g) n/ contans all -factors wth even components. Gven a graph G, we defne an Even Squared Hamltonan Cycle (ESHC) as a Hamltonan cycle C = v v v n v of G wth chords v v +3 for all n. When n 7, an ESHC s 4-regular wth chromatc number χ = for even n and χ = 3 for odd n. It s not hard to check that an n-vertex ESHC contans all bpartte graphs of order n wth maxmum degree at most (e.g., by usng the fact that every ESHC of even order contans a ladder graph defned below). Below s our man result. Theorem.. There exsts N > 0 such that for all even ntegers n N, f G s a graph of order n wth δ(g) n + 9, then G contans an ESHC. We show that the constant 9 n Theorem. cannot be replaced by. Proposton.. Suppose that n 0. Let G be the unon of two copes of K n + sharng 4 vertces. Then δ(g) = n + but G contans no ESHC. We also show that the condton of n beng even s necessary for Theorem. even f we replace 9 by n/8 /. Proposton.3. There are nfntely many odd n and graphs G of order n such that δ(g) n + n/8 / but G contans no ESHC. More generally, an Even kth powered Hamltonan Cycle (EkHC) of a graph G s a Hamltonan cycle v v v n v wth edges v v +j for all n and j k. Then an EHC s smply a Hamltonan cycle whle an EHC s an ESHC. Usng the same proof technques for Theorem., we can derve the followng result, whose proof s omtted. Theorem.4. For any postve nteger k, there exst a constant c = c(k) and a postve nteger N such that f G s a graph of even order n N and δ(g) n + c then G contans an EkHC. One may vew an EkHC of order n = N as the followng bpartte graph. Let B k (N) be the bpartte graph (X Y, E) wth X = {x,..., x N } and Y = { y,..., y N } such that x y j E f and only f j(mod N) { k +,...,, 0,,..., k}. In partcular, B (N), or ESHC, contans the ladder graph defned by Czygrnow and Kerstead [6], whch has the same vertex sets X and Y but x s adjacent to y j f and only f j(mod N) {, 0, }. Note that the ladder graph contans all -factors wth bpartte components. The structure of the paper s as follows. We prove two (easy) Propostons. and.3 n the next secton. Followng the approach of [5] on squared Hamltonan cycles, we prove Theorem. by the regularty method. In Secton 3 we state the Regularty Lemma and the Blow-up Lemma. In Secton 4 we prove Theorem. by provng the non-extremal case and two extremal cases separately. It seems harder to handle the extremal cases here than n [5]; ths s also the reason why we need a large constant 9 n Theorem.. The last secton gathers open problems wth a remark.. Proofs of Propostons. and.3 Gven a graph G, a par (A, B) of vertex subsets s called a separator of G f V(G) = A B, both A B and B A are non-empty and E(A B, B A) =. It s easy to see that Proposton. follows from the followng clam, whch can be proved by a smple case analyss. Clam.. Suppose that G s a graph wth an ESHC. If (A, B) s a separator of G wth A B 3 and B A 3, then A B 6. Proof of Clam.. Let H be an ESHC of G wth Hamltonan cycle C. Assgn C an orentaton. A segment P = xp z of C s called an AB-path f x A B and z B A. Now let x P z be an AB-path such that V(P ) {x, z } A B (ths can be done by lettng P be mnmal). Snce A B 3, B A 3, there s an AB-path x P z contaned n C V(P ) such that V(P ) {x, z } A B. If e(p ) 4 for =,, then A B e(p ) + e(p ) 6 and we are done. On the other hand, we know that e(p ) {, 3} for =, because x z E(G), whch follows from e(a B, B A) =. Wthout loss of generalty, assume that e(p ) =, or P = x y z. By followng the orentaton of C, let x be the predecessor of x and z + be the successor of z. Snce x z, x z + E(G), we have x, z+ A B (see Fg. ).

3 8 G. Chen et al. / Dscrete Mathematcs 3 (0) 6 40 Fg.. A segment connectng A and B. Snce P and P are vertex dsjont AB-paths, we have V(P ) {x, y, z + } =. If P contans at least three nternal vertces, then A B 6 and we are done. So we may assume that P = x y z, and consequently {x, y, z + } A B. If z + x and z+ x, then all x, y, z +, x, y, z + are dstnct, and consequently A B 6. Otherwse, wthout loss of generalty, assume that z + = x. Then P = P z + P s an AB-path wth two vertces n each of A B and B A. Snce A B 3 and B A 3, there s an AB-path x 3 P 3 z 3 whch s vertex-dsjont from P such that V(P 3) {x 3, z 3 } A B. If e(p 3 ) 4, then A B 6 because P, P 3 are dsjont and P contans three vertces y, z +, y from A B. Otherwse P 3 = x 3 y 3 z 3 for some y 3 A B, then {x, 3 y 3, z + } 3 A B. Snce sx vertces y, z +, y, x, 3 y 3, z + 3 are contaned n A B, we have A B 6. Proof of Proposton.3. Let q be an odd prme power, by usng projectve planes, one can construct (e.g., []) C 4 -free graphs H of order h = q + q + wth δ(h) q. Let G := H + K h q,.e., a graph obtaned from H by addng h q vertces such that each new vertex s adjacent to all vertces of H. Let X := V(G) V(H) and n := V(G) = h q. Then n s odd and δ(g) h = n + q n + n 8 because n = q + q + < (q + ). To see that G does not have any ESHC, consder an arbtrary Hamltonan cycle C of G (f t exsts). Snce X s an ndependent vertex set and n s odd, we conclude that C X s the unon of vertex-dsjont paths such that e(c X) s odd. In partcular, one path P[x, y] of C X has odd length. If V(P) 4, then xx +++ E(G), where x +++ = ((x + ) + ) +, snce H contans no C 4. Otherwse V(P) =, whch n turn shows that x, y + X. Snce X s ndependent, x y + E(G). In all cases G does not have an ESHC based on C. 3. The Regularty Lemma and Blow-up Lemma As n [5,8], the Regularty Lemma of Szemeréd [4] and Blow-up Lemma of Komlós et al. [6] are man tools n our proof of Theorem.. For any two dsjont vertex-sets A and B of a graph G, the densty of A and B s the rato d(a, B) := e(a, B)/( A B ), where e(a, B) s the number of edges wth one end vertex n A and the other n B. Let ε and δ be two postve real numbers. The par (A, B) s called ε-regular f for every X A and Y B satsfyng X > ε A, Y > ε B, we have d(x, Y) d(a, B) < ε. Moreover, the par (A, B) s called (ε, δ)-super-regular f (A, B) s ε-regular and deg B (a) > δ B for all a A and deg A (b) > δ A for all b B. Lemma 3. (Regularty Lemma Degree Form). For every ε > 0 there s an M = M(ε) such that, for any graph G = (V, E) and any real number d [0, ], there s a partton of the vertex set V nto l + clusters V 0, V,..., V l, and there s a subgraph G of G wth the followng propertes: l M, V ε V for 0 l, V = V = = V l, deg G (v) > deg G (v) (d + ε) V for all v V, G [V ] = (.e. V s an ndependent set n G ), for all, each par (V, V j ), < j l, s ε-regular wth d(v, V j ) = 0 or d(v, V j ) d n G. The Blow-up Lemma allows us to regard a super-regular par as a complete bpartte graph when embeddng a graph wth bounded degree. We need a bpartte verson of ths lemma whch also restrcts the mappngs of a small number of vertces. Lemma 3. (Blow-Up Lemma Bpartte Verson). For every δ,, c > 0, there exsts an ε = ε(δ,, c) > 0 and α = α(δ,, c) > 0 such that the followng holds. Let (X, Y) be an (ε, δ)-super-regular par wth X = Y = N. If a bpartte graph H wth (H) can be embedded n K N,N by a functon φ, then H can be embedded n (X, Y). Moreover, n each φ (X) and φ (Y), fx at most αn specal vertces z, each of whch s equpped wth a subset S z of X or Y of sze at least cn. The embeddng of H nto (X, Y) exsts even f we restrct the mage of z to be S z for all specal vertces z.

4 G. Chen et al. / Dscrete Mathematcs 3 (0) Proof of Theorem. Let V be the vertex set of a graph G of order n for some even n. A partton V V k of V s called balanced f V V j for all j. In partcular, a balanced bpartton V V satsfes V = V = n/. Gven 0 α, we defne two extremal cases wth parameter α as follows. Extremal Case : There exsts a balanced partton of V nto V and V such that the densty d(v, V ) α. Extremal Case : There exsts a balanced partton of V nto V and V such that the densty d(v, V ) α. The followng three theorems deal wth the non-extremal case and two extremal cases separately. Theorem 4.. For every α > 0, there exst β > 0 and a postve nteger n 0 such that the followng holds for every even nteger n n 0. For every graph G of order n wth δ(g) ( β)n, ether G contans an ESHC or G s n one of the extremal cases wth parameter α. Theorem 4.. Suppose that 0 < α and n s a suffcently large even nteger. Let G be a graph on n vertces wth δ(g) n +3. If G s n Extremal Case wth parameter α, then G contans an ESHC. Theorem 4.3. Suppose that 0 < α and n s a suffcently large even nteger. Let G be a graph on n vertces wth δ(g) n + 9. If G s n Extremal Case wth parameter α, then G contans an ESHC. It s easy to see that Theorem. follows from Theorems For ths purpose we can even use a weaker verson of Theorem 4. wth β = 0, but the current Theorem 4. may have other applcatons. If G s n Extremal Case wth parameter α, then there exsts x V such that deg(x) < ( + α)n/ and n turn δ(g) < ( + α)n/. Theorems 4. and 4. together mply the followng remark, whch s a specal case of the theorem of Böttcher et al. [4]. Remark 4.4. For any α > 0, there exsts a postve nteger n 0 such that every graph G of even order n δ(g) ( + α)n contans an ESHC. Ths remark wll be used n the proof of Theorem 4.3. n 0 wth 4.. Non-extremal case In ths secton we prove Theorem 4.. Proof. We fx the followng sequence of parameters 0 < ε d β α < () and specfy ther dependence as the proof proceeds. Actually we let α be the mnmum of the two parameters defned n the extremal cases. Then we choose d β such that they are much smaller than α. Fnally we choose ε = ε(d, 4, d ) 4 followng the defnton of ε n the Blow-up Lemma. Choose n to be suffcently large. In the proof we omt celng and floor functons f they are not crucal. Let G be a graph of order n such that δ(g) ( β)n and G s not n ether of the extremal cases. Applyng the Regularty Lemma (Lemma 3.) to G wth parameters ε and d, we obtan a partton of V(G) nto l + clusters V 0, V,..., V l for some l M = M(ε), and a subgraph G of G wth all descrbed propertes n Lemma 3.. In partcular, for all v V, deg(v) > deg(v) (d + ε)n G G β ε d n β n, provded that ε + d β. On the other hand, e(g ) e(g) (d+ε) n e(g) dn by usng ε < d. We further assume that l = k s even; otherwse we elmnate the last cluster V l by removng all the vertces n ths cluster to V 0. As a result, V 0 εn and ( ε)n ln = kn n. For each par and j wth j l, we wrte V V j f d(v, V j ) d. As n other applcatons of the Regularty Lemma, we consder the reduced graph G r, whose vertex set s {,..., l}, and two vertces and j are adjacent f and only f V V j. From δ(g ) > ( β)n, a standard argument shows that δ(g r) ( β)l. The rest of the proof conssts of the followng fve steps. Step : Show that G r contans a Hamltonan cycle X Y X k Y k. Step : For each k, ntate a connectng ESP (even squared path) P between Y and X (where Y 0 = Y k ) wth two vertces from each Y and X. Step 3: For each k, move at most εn vertces from X Y to V 0 such that the resultng graph G [X Y ] has the mnmum degree at least (d ε)n. ()

5 30 G. Chen et al. / Dscrete Mathematcs 3 (0) 6 40 Step 4: Extend P,..., P k to nclude all the vertces n V 0 and some vertces n V \ V 0 such that X (V(P ) V(P k )) = Y (V(P ) V(P k )) d N for all k. Step 5: Apply the Blow-up Lemma to each (X, Y ) and obtan an ESP consstng of all the remanng vertces of X Y. Concatenatng these ESP s wth P,..., P k, we obtan the desred ESHC of G. We now gve detals of each step. The assumpton that G s not n ether of the extremal cases leads to the followng clam, whch wll be used n Step and Step 4. Clam 4.5. (a) G r contans no ndependent set U of sze at least ( 8β)l. (b) G r contans no two dsjont subsets U, U of sze at least ( 6β)l such that e G r (U, U ) = 0. Proof. (a) Suppose nstead, that G r contans an ndependent set U of sze ( 8β)l. We wll show that G s n the Extremal Case wth parameter α. Let A = U V and B = V(G) A. By (), 9β n 8β Nl = U N = A < β n. For each x A, snce deg G (x, A) deg G (x, A) + (d + ε)n < βn, we have deg G (x, B) > ( β)n βn ( β)n. Hence e G (A, B) ( 9β)n( β)n > ( 4 β)n. Now move at most 9βn vertces from B to A such that A and B are of sze n/. We stll have e G (A, B) > 4 β n 9βn n 4 n = 0β n = ( 40β). By specalzng 40β α n (), we see that G s n the Extremal Case wth parameter α. (b) Suppose nstead, that G r contans two dsjont subsets U, U of sze ( 6β)l such that e G r (U, U ) = 0. We wll show that G s n the Extremal Case wth parameter α. Let A = U V and B = U V. Snce e Gr (U, U ) = 0, we have e G (A, B) = 0. Snce e(g) e(g ) + dn, we have e G (A, B) e G (A, B) + dn = dn. Note that A = U N = ( 6β)lN > ( 7β)n. Smlarly, B > ( 7β)n. By addng at most 7βn vertces to each of A and B, we obtan two subsets of sze n/ and stll name them as A and B, respectvely. Then, e(a, B) dn + (7βn)(n/) = 8βn, whch n turn shows the densty d(a, B) = e(a, B)/( n ) 3β. Snce α > 3β, we obtan that G s n the Extremal Case wth parameter α. Step. To show that G r s Hamltonan, we need the followng theorem of Nash-Wllams. Theorem 4.6 (Nash-Wllams []). Let G be a -connected graph of order n. If mnmum degree δ(g) max{(n + )/3, α(g)}, then G contans a Hamltonan cycle. We frst show that G r s βl-connected. Suppose, to the contrary, let S be a cut of G r such that S < βl and let U and U be two components of G r S. Snce δ(g r ) ( β)l, we have U ( 3β)l for =,. Snce e(u, U ) = 0, we obtan a contradcton to Clam 4.5(b). Snce n = Nl + V 0 (l + )εn, we have l /ε 3/β, provded that β 5ε. Then βl 3, and G r s 3-connected. By Clam 4.5(a), we have α(g) ( 8β)l < δ(g r). By Theorem 4.6, G r s Hamltonan. Followng the order of a Hamltonan cycle of G r, we denote all the clusters of G except for V 0 by X, Y,..., X k, Y k (recall that l = k s even). We call X, Y partners of each other and wrte P(X ) = Y and P(Y ) = X. Step. For each =,..., k, we ntate an ESP P of G connectng X and Y (wth Y 0 = Y k ) as follows. Gven an ε-regular par (X, Y) of clusters and a subset Y Y, we call a vertex x X typcal to Y f deg(x, Y ) (d ε) Y. By the regularty of (X, Y), at most εn vertces of X are not typcal to Y whenever Y > εn. Fx k. Frst let a X be a vertex typcal to both Y and Y and let b X be a vertex typcal to both Γ (a, Y ) and Γ (a, Y ). Snce both pars (Y, X ) and (X, Y ) are ε-regularty wth densty at least d, all but εn vertces of X can be chosen as a. Snce Γ (a, Y ), Γ (a, Y ) (d ε)n > εn, all but at most εn + vertces of X can be chosen as b (note that b a ). Recall that Γ (a b, Y ) = Γ (a, Y ) Γ (b, Y ). The way we select a and b guarantees that Γ (a b, Y ) (d ε) N εn +. Now let c, d Γ (a b, Y ) be two (dstnct) vertces of Y such that c s typcal to both X and X, and d s typcal to both Γ (c, X ) and Γ (c, X ). All but at most εn vertces of Γ (a b, Y ) can be chosen as c and d. In summary P = c a d b s an ESP wth c, d Y, a, b X such that deg(c d, X ) (d ε) N, deg(a, Y ) (d ε)n, deg(a b, Y ) (d ε) N, deg(d, X ) (d ε)n. (3)

6 G. Chen et al. / Dscrete Mathematcs 3 (0) Step 3. For each, let X := {x X, deg(x, Y ) (d ε)n} and := { y Y, deg( y, X ) (d ε)n}. Y Snce (X, Y ) s ε-regular, we have X, Y ( ε)n. If X Y, say X > Y, then we pck an arbtrary subset of X of sze Y and stll name t X. As a result, we have X = Y. Let V := 0 V 0 k = (X X ) (Y Y ). From X X + Y Y εn, we derve that V 0 εn+(εn)k = 3εn by usng Nk n from (). In addton, the mnmum degree δ(g[x, Y ]) (d ε)n εn. It s easy to see that (X, Y ) s ε-regular [9, Slcng Lemma]. Step 4. Consder a vertex x V(G) and an orgnal cluster A (X or Y for some ), we say that x s adjacent to A, denoted by x A, f deg(x, A) (d ε)n. Gven two vertces u, w, we defne a u, w-chan of length t as dstnct clusters A, B,..., A t, B t such that u A B A t B t w and each A j and B j are partners, n other words, {A j, B j } = {X j, Y j } for some j. Clam 4.7. Let L be a lst of at most εn pars of vertces of G. For each {u, w} L, we can fnd u, w-chans of length at most four such that every cluster s used n at most d N/0 chans. Proof. Suppose that we have found chans of length at most four for the frst m < εn pars such that no cluster s contaned n more than d N/0 chans. Let Ω be the set of all clusters that are used n exactly d N/0 chans. Snce each chan uses at most four clusters, we have d 0 N Ω 4m 8εn 8ε kn ε, where the last nequalty follows from (). Therefore Ω k 30ε l βl provded that ε and ( ε)d d 30ε d β. Now consder a par {u, w} L. Our goal s to fnd a u, w-chan of length at most four by usng clusters not n Ω. Let U be the set of all clusters adjacent to u but not n Ω, and W be the set of all clusters adjacent to w but not n Ω. Let P(U) and P(W) be the set of the partners of clusters n U and W, respectvely. The defnton of chans mples that a cluster A Ω f and only f ts partner P(A) s n Ω. Therefore (P(U) P(W)) Ω =. We clam that P(U) = U ( 3β)l. To see t, we frst observe that any vertex v V s adjacent to at least ( β)l clusters. For nstead, β n deg(v) G β ln + dnl + 3εn < 3 β n, a contradcton, provded that β d + 3ε. Snce Ω βl, we thus have U ( 3β)l. Smlarly P(W) = W ( 3β)l. If E Gr (P(U), P(W)), then there exst two adjacent clusters B P(U), A P(W). If B, A are partners of each other, then u A B w gves a u, w-chan of length two. Otherwse assume that A = P(B ) and B = P(A ). Then u A B A B w gves a u, w-chan of length four. Note that all A, B Ω. We may thus assume that E Gr (P(U), P(W)) =. If P(U) P(W) =, then (4) contradcts wth Clam 4.5(b) because P(U) ( 3β)l and P(W) ( 3β)l. Otherwse assume that A P(U) P(W). Then by (4), A s not adjacent to any cluster n P(U) P(W). Snce deg Gr (A) ( β)l, we derve that P(U) P(W) ( + β)l. Snce P(U) ( 3β)l and P(W) ( 3β)l, then P(U) P(W) ( 8β)l. By (4), P(U) P(W) s an ndependent set n G r, whch contradcts wth Clam 4.5(a). We arbtrarly partton V 0 nto at most εn pars (note that V 0 s even because X = Y for all ). Applyng Clam 4.7, we construct chans of length at most four for each par such that every cluster s used n at most d N/0 chans. For each let m denote the number of chans contanng X and Y. Clam 4.8. We can extend connectng ESP s to nclude all the vertces n V 0 such that the followng holds for all. The resultng ESP s P = u v u t v t satsfes u, u Y, v t, v t X and 30ε (4) deg(u u, X ) (d ε) N, deg(v t v t, Y ) (d ε) N, deg(v, Y ) (d ε)n, deg(u t, X ) (d ε)n. The sets X = X j V(P j ) and Y = Y j V(P j ) satsfy X = Y ( ε)n 7m. Proof. We prove by nducton on m := V 0 /. When m = 0, by (3), the ntal P X = X {a, b } and Y := Y {c, d }) satsfy (6). (5) (6) = c a d b satsfes (5). The ntal

7 3 G. Chen et al. / Dscrete Mathematcs 3 (0) 6 40 Suppose that m and we have extended the connectng ESP s to nclude m pars from V 0 such that (5) and (6) hold for all. Let {x, y} be the last par from V 0. We frst consder the case when the x, y-chan has length two. Wthout loss of generalty, assume that x Y X y for some. Let P = u v u t v t be the current connectng ESP between Y and X and Y := Y j V(P j ) and X := X j V(P j ). To nclude x, we extend P to P = P y x y x y 3 x 3 y 4 x wth four vertces y, y, y 3, y 4 Y and three vertces x, x, x 3 X y Γ (v t v t ), y Γ (v t ), y 3, y 4 Γ (x), x Γ (u t ), such that n addton y 4 s typcal to X, and x 3 s typcal to Γ (x, Y ). (7) Ths s possble by usng Lemma 3. (actually we only need the regularty between X and Y ; but applyng the Blow-up Lemma makes our proof shorter). To see t, frst note that (6) mples that X, Y > ( d /)N because m d N/0 by Clam 4.7. Then, by (5), Γ (v t v t, Y ) deg(v t v t, Y ) d N (d ε) N d N > d 4 N. Smlarly we can show that Γ (v t, Y ), Γ (u t, X ) (d d )N. The defnton of x Y also guarantees that Γ (x, Y ) (d d )N. Fnally (7) only forbds addtonal εn vertces when choosng y 4 and x 3. Therefore we can apply Lemma 3. to fnd such an P. By (7), we have deg( y 4, X ) (d ε)n and deg(x 3 x, Y ) (d ε) N. Consequently P satsfes (5). Snce x behaves lke a vertex of X n P, we call such a procedure nsertng x nto X (by extendng P ). Snce y X, we can smlarly nsert y to Y by extendng P + to P = + yx 4y 5 x 5 y 6 x 6 y 7 x 7 P + wth x j X and y j Y. Snce each X and Y lose seven vertces totally, (6) holds. Now consder the case when the x, y-chan has length four. Assume that x A B A B y. We frst nsert x to B, then pck any (avalable) vertex n B that s typcal to A and nsert t to B, and fnally nsert y to A. As a result, A, B each loses four vertces to some connectng paths whle A, B each loses seven vertces to some connectng paths. Thus (6) holds. Step 5. Fx k. Suppose that at present P = u v u t v t and P + = w z w s z s for some ntegers s, t, and X = X j V(P j ) and Y = Y j V(P j ). By Clam 4.8, P and P + satsfy (5) and X, Y ( d /)N. Snce (X, Y ) s ε-regular, the Slcng Lemma of [9] says that (X, Y ) s (ε, d/)-super-regular (note that (d ε)n d N/ > dn/). We now apply the Blow-up Lemma to each G [X, Y ] to obtan an spannng ESP y x y x N, y N, x N (see Fg. ), where N = X = Y such that Fg.. An ESP coverng X and Y. y Γ (v t v t, Y ), x N Γ (w, X ), x Γ (u t, X ), y N Γ (z, Y ), y Γ (v t, Y ), x N Γ (w w, X ). The restrctve mappng of y, x, y, x N, y N, x N s possble because by (5), all the targetng sets are of sze at least (d ε) N d N/ > d N/4. We now complete the proof of the non-extremal case. 4.. Extremal Case In ths subsecton we prove Theorem 4.. We start wth a lemma whch gves a balanced spannng bpartte subgraph that we wll use throughout the secton.

8 G. Chen et al. / Dscrete Mathematcs 3 (0) Lemma 4.9. Suppose that 0 α (/9) 3. Let G = (V, E) be a graph on n vertces wth δ(g) n + 3 and a balanced partton V V such that d(v, V ) α. Then G contans a balanced spannng bpartte subgraph G wth parts U, U such that There s a set W of at most α /3 n vertces such that we can fnd vertex dsjont 4-stars (stars wth four edges) n G wth the vertces of W as ther centers. deg G (x) ( α /3 α /3 )n/ for all x W. Proof. For smplcty, let α = α /3 and α = α /3. For each =,, we defne V = x V : deg(x, V 3 ) ( α ) n. Snce d(v, V ) α, we have V V α n/ and consequently V ( α )n/ for =,. For any x V, deg(x, V ) > 3 ( α ) n α n. Let V 0 = V V V. Then V 0 α n. For each v V 0 and =,, we have deg(v, V ) ( α ) n, whch mples that n deg(v, V ) α and (8) deg(v, V ) (α α ) n. (9) We now separate cases based on V and V. Case : V, V n/. In ths case we partton V 0 nto W W such that W = n/ V for =,. For each vertex w W, we greedly fnd four neghbors from V 3 such that the neghbors for all the vertces of W are dstnct. Ths s possble for =, because of (9) and (α α ) n 4α n 4 V 0 provded that α 9α or α /9. Defne U = V W for =,. Then U = U = n/. Wth W = V 0, the second asserton of Lemma 4.9 follows from (8). Case : one of V, V, say, V s greater than n/. Let V 0 be the set of vertces v V such that deg(v, V ) α n/. Frst assume that V 0 V n/. Then we form a set W wth V n/ vertces of V 0 and all the vertces of V 0. Let U = V W and U = V W. Then U = U = n/. Snce ( α )n/ V n/, we have W α n/. Then V U α n/, by (9) and the defnton of V 0, we have deg(v, U ) deg(v, V ) α n/ (α α )n/ α n/ (0) for all v W. Wth α 6α, we have (α α )n/ 4α n/ 4 W. Therefore we can greedly fnd four neghbors for each vertex v W such that the neghbors for all the vertces of W are dstnct. The second asserton of Lemma 4.9 follows from (8) and V U α n/. Otherwse assume that V 0 < V n/. In ths case let U = V V 0 and U = V V 0 V 0. Then U = n + t for some postve nteger t α n/. Snce deg(v, V ) < α n/ for every v U, the nduced graph G[U ] has the maxmum degree α n/. By the mnmum degree condton δ(g) n/ + 3, G[U ] has the mnmum degree at least n n δ(g) U + 3 t t + 3. We now need the followng smple fact. Fact 4.0. In a graph G of order n wth the maxmum degree (G ) and the mnmum degree δ(g ) t, the number of dsjont 4-stars s at least (t 3)n 5( +t 3). To see t, suppose G has a largest famly of dsjont 4-stars on some vertex set M of sze m. Then (t 3)(n 5m) e(m, V(G) M) 5m and the fact follows. Applyng Fact 4.0, there are at least (t + 3 3) U 5( + t + 3 3) t U n 5 α + t t n 5(α + α ) n t vertex dsjont 4-starts n G[U ]. Pck t such 4-stars and move ther centers to U. As a result, U = U = n/. Let W 0 = V 0 V 0 and W be the unon of W 0 wth the new vertces of U. Snce V U = V n/ α n/, we have (0) for all v W 0. Snce W = n/ V α n/, we can fnd dsjont 4-stars n G[U, W 0 ] wth all the vertces of W 0 as centers such that these 4-stars are also dsjont from the exstng 4-stars. In addton, the second asserton of Lemma 4.9 holds as before.

9 34 G. Chen et al. / Dscrete Mathematcs 3 (0) 6 40 Proposton 4. shows that f G s a graph or a bpartte graph wth a large mnmum degree and t contans not many vertex dsjont 4-stars, then we can fnd an ESC or ESP contanng all the vertces n these stars. We need ts part () for ths subsecton, and part () for Extremal Case. Proposton 4.. Fx 0 < ε /5. (a) Let G be a graph of order N wth a subset W of sze t ε N/8. Suppose that G contans t vertex-dsjont 4-stars wth the vertces of W as centers, and deg(x) ( ε )N for every vertex x W. Then G contans an ESC C of length 8t whch contans all the vertces of W such that any two nearest vertces of W are separated by exactly seven vertces not n W. (b) Let G be a bpartte graph on two parts U, U of sze N. Let W be a vertex subset of sze t ε N/5. Suppose that G contans t vertex-dsjont 4-stars wth the vertces of W as centers, and deg(x) ( ε )N for every vertex x W. Then G contans an ESP of length 8t + 4 whch contans all the vertces of W and whose frst and last three vertces are not from W. Proof. (a) Suppose that W = {w,..., w t }, and denote the four leaves under w by a, b, c, d. For each, we greedly choose three new vertces u, v, x that are not contaned n any exstng star such that u Γ (c d a b ), v Γ (d a b c ), x Γ (b c d a + ), n whch the ndces are modulo t. Ths s possble because each a, b, c, d, t, has at least ( ε )N neghbors and any four of them have at least ( 4ε )N ε N 8t common neghbors (where 8t s the total number of vertces used at the end of ths greedy algorthm). We thus obtan an ESC u a v b w c x d : =,..., t, n whch the vertces of W are dstrbuted evenly. (b) Partton W = W W wth W = U W and W = U W. For each W, we follow the procedure n () to fnd two vertex dsjont ESC s C and C of length 8 W and 8 W n whch the vertces of W are dstrbuted evenly. The calculaton s smlar except that any four vertces n U W (or U W ) have at least ( 4ε )N ε N > 4t common neghbors n U (or U ), where 4t s the total number of the vertces used n one partton set. We next break C nto P = x x x 3 u 3 u u and break C nto P = v v v 3 y 3 y y such that x, u, v, y W for =,, 3 and u, u 3, v U, u, v, v 3 U. Assume that t otherwse we are done. Choose four new vertces not n W (n ths order) z Γ (u u 3 ), z 3 Γ (u v ) and z Γ (u z z 3 v ), z 4 Γ (z z 3 v v 3 ). Ths s possble because the number of common neghbors of any four vertces not n W s at least ( 4ε )N 5t 4t +, where 4t + s the total number of the vertces used n one partton set. As a result, P z z z 3 z 4 P s an ESP whch contans all the vertces of W and whose frst and last three vertces are not from W. We fnally observe that a bpartte graph wth very large mnmum degree s super-regular. Proposton 4.. Gven 0 < ρ <, let G be a bpartte graph on X Y such that δ(x, Y) ( ρ) Y, δ(y, X) ( ρ) X. () Then G s ( ρ, ρ)-super-regular. Proof. It suffces to show that G s ρ-regular. Consder subsets A X, B Y wth A = ε X and B = ε Y for some ε, ε > ρ. By (), we have δ(a, Y) Y ρ Y and consequently δ(a, B) B ρ Y = (ε ρ) Y. The densty between A and B satsfes d(a, B) δ(a, B) A A B (ε ρ) Y B = ε ρ ε > ρ ρ = ρ. Snce ρ < d(a, B) and n partcular, ρ < d(x, Y), we have d(a, B) d(x, Y) < ρ. We are ready to prove Theorem 4. now. Proof of Theorem 4.. Let 0 α, n partcular α (/9) 3. Wrte α = α /3 and α = α /3. Let G = (V, E) be a graph on n vertces wth δ(g) n + 3. Suppose G s n Extremal Case wth parameter α. We frst apply Lemma 4.9 to G and obtan a bpartte subgraph G wth two partton sets U, U of sze n/ whch contans at most α /3 n vertex dsjont 4-stars. Denote by W the set of the centers of the 4-stars. We also have deg G (x) ( α α )n/ for all x W. () Snce α + α /5 and α n α +α 5 n, we may apply Proposton 4.(b) to G wth ε = α + α and N = n/. We thus obtan an ESP P 0 = x x x 3 y 3 y y of length 8 W + 4 whch contans all the vertces of W such that x, y V V for 3. In order to fnd an ESHC of G, t suffces to fnd an ESP P = u u u 3 v 3 v v on V(G) V(P 0 ) such that x 3 x x Py y y 3 = x 3 x x u u u 3 v 3 v v y y y 3 s also an ESP.

10 G. Chen et al. / Dscrete Mathematcs 3 (0) Let U = U V(P 0 ) for =,, and n = U = U. Then n = n/ (4 V 0 + ) n/ (4α n + ). By (), the bpartte subgraph G[U, U ] has ts mnmum degree at least ( α α ) n (4α n + ) ( 3α ) n ( 3α )n by usng α 9α. Smlarly the degree from x or y, =,, 3, to U or U s at least ( 3α )n. By Proposton 4., (U, U ) s ( 3α, )-super-regular (usng α 3 /9 agan). Snce 3α, we can apply the Blow-up Lemma to obtan an ESHP u u u 3 v 3 v v of G[U, U ] such that u Γ (x, x 3 ), u Γ (x ), u 3 Γ (x ), v Γ ( y, y 3 ), v Γ ( y ), v 3 Γ ( y ). Snce Γ (x ), Γ ( y ) ( 3α )n for =,, 3 and Γ (x, x 3 ), Γ ( y, y 3 ) ( 6α )n, the restrctve mappng of u, u, u 3 and v, v, v 3 s possble Extremal Case In ths subsecton we prove Theorem 4.3. In Extremal Case, we used the Blow-up Lemma to fnd an ESHP wth certan propertes n a bpartte graph wth very large mnmum degree. In ths subsecton we frst prove such a lemma for arbtrary graphs. Lemma 4.3. Let k 3 and n be suffcently large. Let G be a graph of order k + n + 6. Suppose that G contans an ESP P 0 = u u k. Let X = V(G) V(P 0 ). Suppose that x x x 3 and y y y 3 are two paths n X and let X be the set of the remanng vertces of X (then X = n ). If deg(x, X ) 7 n 8 + for all vertces x X and deg(u j, X) 7 X + for 8 j =,, 3, k, k, k, then G contans an ESHP that starts wth x 3 x x, fnshes wth y y y 3 and contans P 0 as an nternal path. Proof. Our proof conssts of three steps. Step : we fnd an ESC on X. Let G = G[X ]. Snce δ(g ) 7 n 8 and n s suffcently large, by Remark 4.4, G contans an ESHC. Step : we fnd an ESP on X such that t starts wth x 3 x x and fnshes wth y y y 3. Let v,..., v n be the ESC gven by Step. We wll form an ESP x 3 x x v v v n v n v + y y y 3 for some n. It suffces to have the followng adjacences. x 3 v, x v, x v, x v, y v +, y v +3, y v +, y 3 v +, n whch we assume that v j = v j+n for all ntegers j. Snce deg(x, X ) 7n /8 + for any vertex x {x, x, x 3, y, y, y 3 }, the number of n satsfyng (3) s at least n 8(n /8 ) = 8. Thus (3) holds for some n. Step 3: we fnd an ESHP of G that starts wth x 3 x x, fnshes wth y y y 3 and contans P 0 as an nternal path. Let n = X = n + 6. Denote the ESP found n Step by v,..., v n, where v = x 3, v = x, v 3 = x, v n = y, v n = y, v n = y 3. Our goal s to fnd an ndex 3 n 3 such that v v P 0 v + v n s an ESP. Snce P 0 = u u k, t suffces to have the followng adjacences. u v, u v, u 3 v, u v, u k v +, u k v +3, u k v +, u k v +, for some 3 n 3. Snce deg(u j, X) 7n /8 + for j =,, 3, k, k, k, the number of n satsfyng (3) s at least n 8(n /8 ) = 8. Thus (3) holds for some 3 n 3. Proof of Theorem 4.3. We start wth defnng two new sets, whch are varants of V and V. Let α = α /3 and α = α /3. We defne V n = x V : deg(x, V 3 ) < α for =,. Snce δ(g) > n/, we have deg(x, V ) > ( α )n/ for every x V. Snce d(v, V ) α, we have V V α n/ and V ( α )n/ for =,. Consequently, deg(x, V ) > deg(x, V n ) α ( α α ) n for all x V. (5) Let V 0 = V V V. Then V 0 α n and deg(x, V ) (α α )n/ for all x V 0. (3) (4)

11 36 G. Chen et al. / Dscrete Mathematcs 3 (0) 6 40 Our proof conssts of the followng two steps whch together provde an ESHC of G. Step. Fnd two dsjont ESP s x x p and y y p of length 6 p 4 such that x, x, x 3, y, y, y 3 V and x p, x p, x p, y p, y p, y p V. Step. Fnd two ESP s P and P consstng of all the remanng vertces n V and V V 0, respectvely, such that x 3 x x P y y y 3, and x p x p x p P y p y p y p are also ESP s. Whle Step follows from Proposton 4. and Lemma 4.3 easly, Step s much harder (at least from our pont of vew) t s where we need the large constant 9 n the mn-degree condton. Below we present Step frst Step : Fnd two ESP s coverng the remanng vertces Let P = x x p and P = y y p be the two ESP s of length 6 p 4 provded by Step. Let S = V(P ) V(P ). Let U = V S and U = U {x, x, x 3, y, y, y 3 }. Fx x U, by (5), we have deg(x, U ) ( α α )n/ p ( α α )n/ +, (6) where the second nequalty follows from α n/ 9 p +. Usng U ( + α )n/ and α 3α, we derve that deg(x, U ) ( α ) U +. Usng α, we have deg(x, 6 U ) 7 8 U +. We then apply Lemma 4.3 to G[U ] wth P 0 = and obtan an ESP x 3 x x y y y 3 on U. Let U = V V 0 S and U = U {x p, x p, x p, y p, y p, y p }. Partton U nto V = V S and V = 0 V 0 S. We have V α 0 n and for each x V 0, deg(x, V ) (α α ) n p α n 4 V, 0 by usng α n/ p and α 6α. We then greedly fnd V 0 dsjont 4-stars wth the vertces of V 0 as centers and vertces n V as leaves. Let N = ( + α )n/. Then U V V 0 = n V N. For any vertex x V, the arguments above for G[U ] gve that deg(x, V ) ( α )N +. (7) We then apply Proposton 4.(a) to G[U ] and obtan an ESC C 0 of length k = 8 V 8α 0 n such that t contans all the vertces of V 0 n a way that every two nearest vertces of V 0 are separated by exactly seven vertces of V. We then break C 0 nto an ESP P 0 = u u k such that u, u, u 3, u k, u k, u k V. Let X = U V(P 0) and X = X {x p, x p, x p, y p, y p, y p }. For any vertex x V, by (7) and usng n 8α n α α N, deg(x, X ) ( α )N + 8α n ( 3α )N +. Snce X V, u, u, u 3, u k, u k, u k V and X < X N, by lettng α, the degree condtons n Lemma hold. We then apply Lemma 4.3 to G[U ] and obtan the desred ESHP x p x p x p P 0 y p y p y p Step : Connect V and V Gven two dsjont sets A and B, an ESP on A B s called an (A, B)-connector f ts frst three vertces are from A, and the last three vertces are from B. Our goal s to fnd two dsjont (V, V )-connectors of length at most 4. The smplest connector s an ESP x x 6 wth x, x, x 3 V and x 4, x 5, x 6 V. Unfortunately such a smple connector may not exst f e(v, V ) s very small but e(v, V 0) s relatvely large. Let us sketch our proof. We frst separate the vertces of V 0 wth large degree to both V and V : V 0 = {x V 0 : deg(x, V ), deg(x, V ) n/6 α n}. The reason why we choose n/6 can be seen from (0), n whch we use n/ = 3(n/6). If V 0 > 65, then we can fnd two dsjont copes of T,3, from the 3-partte subgraph G[V, V, 0 V ], where T,3, s the unon of two copes K,3 sharng the three vertces n one partton set. Each copy of T,3, can be easly extended to an (V, V )-connector. If V 0 65, then V 0 wll not be used any more. We add the vertces of V 0 V 0 nto V or V formng two new (dsjont) sets U, U such that any three vertces n U, =,, have many common neghbors. What remans s to fnd two dsjont (U, U )-connectors by usng the mnmum degree condton δ(g) (n + V 0 )/ + 9. One way to construct such a connector s to fnd two adjacent vertces x U, y U such that there s 4-vertex path between Γ (x, U ) and Γ ( y, U ). If ths cannot be done, then we fnd sx vertces x, x, x 3 U and x 4, x 5, x 6 U such that x x 4 and x 3 x 6 are edges and x, x 3, x 4, x 5 form a copy of K, n G[U, U ]. After fndng one connector, we remove all or some of ts vertces and repeat the procedure above. Note that gnorng V 0 of sze at most 65 s the major reason for the large constant 9 n δ(g); for example, when V 0 =, then δ(g) n/ + 9 suffces. We need the followng propostons on the exstence of T,3, and K,, whose proofs are standard countng arguments. Gven two dsjont vertex sets A, B n a graph H, we wrte K s,t (A, B) f H[A, B] contans a copy of K s,t wth s vertces from A and t vertces from B. Smlarly T,3, (A, B, C) means that there are subsets X A, Y B and Z C such that H[X, Y] = H[Z, Y] = K,3. We denote by δ(a, B) the mnmum degree deg(a, B) over all a A.

12 G. Chen et al. / Dscrete Mathematcs 3 (0) Proposton 4.4. (a) Gven any nteger t > 0, there exst ε 0 > 0 and n 0 such that the followng holds for any ε ε 0 and n n 0. Let A, B be two dsjont vertex sets n a graph such that δ(b, A) n/6 3εn, A ( + ε)n/ and B > 9(t ). Then K,t (A, B). (b) There exst ε 0 > 0 and n 0 such that the followng holds for any ε ε 0 and n n 0. Let A, B, C be three dsjont vertex set n a graph such that δ(b, A), δ(b, C) n/6 3εn, A, C ( + ε)n/ and B > 6. Then T,3, (A, B, C). Proof. (a) If the graph contans no K,t wth vertces n A and t vertces n B, then deg(x, A) A (t ). x B (8) Snce δ(b, A) n/6 3εn and A ( + ε)n/, n/6 3εn ( + ε)n/ B (t ). As n and ε 0, we obtan B 9(t ), contradcton. (b) Snce B > 9(9 ), we can apply () wth t = 9 to (A, B) and obtan a copy of K,9 on X A of sze and Y B of sze 9. Then snce Y = 9 > 9(3 ), we can apply () agan wth t = 3 to (C, Y ) and obtan a copy of K,3 on Z C of sze and Y Y of sze 3. The next proposton easly follows from a classcal result of Kővár et al. [0]. Proposton 4.5. Let H = (A B, E) be a bpartte graph such that A = n, B = m. Then H contans a copy of K, f ether of the followng holds. (a) deg(x) m for all x A and n > m + m, (b) e := E max{3n, m /}. Now we start our proof. Frst assume that V > Snce δ(v, 0 V ) n/6 α n and V (+α )n/, we can apply Proposton 4.4(b) to the 3-partte subgraph on V V 0 V and fnd a copy of T,3, on X V, Y V 0 and Z V such that X = Z =, Y = 3. Let V = V X, V = 0 V 0 Y, and V = V Z. Then V 0 > 6 and δ(v, 0 V ) n/6 α n. We apply Proposton 4.4(b) agan to 3-partte subgraph on V V 0 V and fnd another copy of T,3,. We next extend each copy of T,3, to a (V, V )-connector of length as follows. Assume X = {x 3, x 5 }, Y = {x 4, x 6, x 8 }, and Z = {x 7, x 9 }. Then x 3, x 4,..., x 8, x 9 s an ESP but t s not a (V, V )-connector because x 4 V and x 8 V. We extend ths ESP by addng two vertces from V n the begnnng and two vertces from V at the end. Snce x 3, x 5 V, by (5), we can fnd a vertex x Γ (x 3 x 5, V ). Snce deg(x, V ) > V α n/ and deg(x 4, V ) > n/6 α n, we can fnd a vertex x Γ (x x 4, V ), whch s dfferent from x 3, x 5. Therefore x x x 9 s an ESP. Smlarly we fnd x 0, x V such that x x,..., x 0 x s an ESP, whch s a (V, V )-connector. Now assume that c 0 := V We wll not use the vertces of V 0 any more. Snce V 0 α n, all vertces x V 0 satsfy deg(x, V V ) > n/ α n. If x V 0 V 0, then exactly one of deg(x, V ) and deg(x, V ) s less than n/6 α n. We thus partton V 0 V 0 nto W and W such that W = {x V 0 V : deg(x, 0 V ) < 3 n/6 α n}. For =,, we have δ(w, V ) δ(w, V V ) n 6 + α n n α n n 6 + α n = n 3 + α n. Let U = V W for =,. The above bound for δ(w, V ) and (5) together mply that δ(u, V ) n 3 + α n. (9) Snce V ( + α )n/, for any three vertces x, x, x 3 U, we have deg(x x x 3, V ) 3 n 3 + α n V α n. (0) Wthout loss of generalty, assume that U U. Snce U ( α )n/, we have U (+α )n/. It suffces to fnd two dsjont (U, U )-connectors u u q and v v q for some q 8. In fact, f any of u, u, u 3 s not from V (note that u, u, u 3 U by the defnton of (U, U )-connectors), then we fnd at most three new vertces x, x, x 3 V such that x x x 3 u u u 3 s an ESP. For example, assume that u V. Then we frst fnd x 3 Γ (u u 3, V ), then x Γ (x 3 u, V ), and fnally x Γ (x u, V ) by applyng (0) three tmes. Smlar we add at most three new vertces x q+, x q+, x q+3 from V such that u q u q u q x q+ x q+ x q+3 s an ESP. The resultng ESP x x x 3 u u q x q+ x q+ x q+3 s a (V, V )-connector of length at most 4. We obtan the other connector smlarly.

13 38 G. Chen et al. / Dscrete Mathematcs 3 (0) 6 40 The followng techncal lemma s the man step n our proof, we postpone ts proof to the end. Lemma 4.6. Let ε and n be suffcently large. Suppose that G s a graph of order n wth a vertex partton U 0 U U such that U 0 = c 0 n; U U ( + ε)n/; U contans a subset V such that δ(v, V ) ( ε)n/; δ(u, U ) n/3. If δ(g) n+c 0 + 6, then G[U, U ] contan ether of the followng 6-vertex subgraphs. H : Two vertces x U, x 5 U are adjacent; two vertces x, x 3 Γ (x, V ) and two vertces x 4, x 6 Γ (x 5, U ) form a path x x 4 x 3 x 6. H : Four vertces x, x 3 U and x 4, x 5 U form a copy of K, ; a vertex x Γ (x 4, U ), and a vertex x 6 Γ (x 3, U ). We observe that subgraphs H and H gven by Lemma 4.6 can be easly converted to (U, U )-connectors. In the case of H, x x x 3 x 4 x 5 x 6 s an ESP and thus a (U, U )-connector. In the case of H, by (0), x, x, x 3 have a common neghbor x 7 V, and x 4, x 5, x 6 have a common neghbor x 8 V. The x x 7 x x 4 x 3 x 5 x 8 x 6 s an ESP and thus a (U, U )-connector (see Fg. 3). Fg. 3. Convert H and H to ESP s. Recall that δ(g) n/ + 9 (n + V 0 )/ + 9. Wth ε = α + α, the partton V 0 U U satsfes the condton of Lemma 4.6 because of (5) and (9). Applyng Lemma 4.6, we obtan ether H or H wth vertex set {x,..., x 6 }. Now let V = 0 V 0 {x,..., x 6 }, U = U {x, x, x 3 }, and U = U {x 4, x 5, x 6 }. Then δ(g) (n + V 0 )/ + 6 and the new partton V 0 U U stll satsfes the condton of Lemma 4.6. By Lemma 4.6, we can fnd a copy of H or H from G[U, U ]. We fnally convert the two copes of H or H to two dsjont (U, U )-connectors. Ths complete the proof of Theorem 4.3 and the man theorem. Proof of Lemma 4.6. Defne k := c 0 / + 6. By assumpton δ(g) n/ + k. Let U = n/ b. Snce U (n c 0 )/, we have b c 0 /. δ(u, U ) n n + k b c 0 = k + b c 0 k c 0 = 6. () However, we do not have a nontrval lower bound for δ(u, U ) because t may be the case that k b. Defne U = {u U : u x for some x V }. Note that U because of (). Select x 5 U such that deg(x 5, U ) = max u U deg(u, U ). Pck an arbtrary vertex x Γ (x 5, V ). Let B = Γ (x, V ) and B = Γ (x 5, U ). By assumptons, B ( ε)n/ and B n/3. If the bpartte graph G[B, B ] contans a 4-vertex path x x 4 x 3 x 6, then we mmedately obtan the desred graph H. We may therefore assume that G[B, B ] contans no 4-vertex path. Ths mples G[B, B ] conssts of dsjont stars, n partcular, e(b, B ) < B + B. Let B = {x B : deg(x, B ) }. The vertces n B B thus have dsjont neghborhoods n B of sze at least. Consequently B B B / U / ( + ε)n/4. Therefore B n/4 εn (n partcular B ). Let A = U B and set m = A. Snce B n/3, then m (+ε)n/ n/3 n/6+εn/. By () and the defnton of B, we have δ(b, A ) k + b c 0. In partcular, m k+b c 0. On the other hand, the defnton of x 5 says that deg(u, U ) deg(x 5, U ) = n/+b c 0 m for every u U. By usng m k + b c 0, we obtan δ(u, U ) n n + k + b c 0 m c 0 = k b + m (3) () k c 0 =. (4)

14 G. Chen et al. / Dscrete Mathematcs 3 (0) We observe that t suffces to fnd a copy of K, from G[U, U ]. In fact, assume that x, x 3 U and x 4, x 5 U are the four vertces of K,. By (), x 3 has a neghbor x 6 n U {x 4, x 5 }; by (4), x 4 has a neghbor x U {x, x 3 }. Ths gves the desred graph H. We now separate cases by whether b m + c 0. Case : b m + c 0. By (), δ(b, A ) k + m + c 0 c 0 > m as k > c 0 / +. Snce B n/4 εn and m n/6 εn/, we have B > A + A. By Proposton 4.5, G[B, A ] contans a copy of K,. Note that the two vertces of ths K, n A belong to U because they have neghbors n B V. Case : b < m + c 0. Let A = U A. By (3), δ(a, U ) > k ( m + c 0 ) + m + m m m/, where the last nequalty holds for any m 0. Ths mples e(a, U ) A /. On the other hand, () mples that e(b, U ) 6 B because k c 0 6. Recall that e(b, B ) < B + B. By the defnton of A, e(b, A ) = e(b, A ) > 6 B ( B + B ) > 5( ε) n ( + ε) n > 3n. Consequently e(u, A ) e(b, A ) > 3n/ 3 U. We can apply Proposton 4.5 and obtan a copy of K, n G[U, A ]. 5. Open problems What s the smallest nteger C such that every graph of even order n wth δ(g) n/+c contans an ESHC? Theorem. and Proposton. together show that C 9. We thnk the lower bound s closer to the truth. To justfy t, one needs to mprove the constants n Theorems 4. and 4.3. What s the mnmum degree threshold δ(n) for ESHC s of odd order? The (general) theorem of Böttcher, Schacht and Taraz (see the footnote on page ) mples that δ(n) n/ + εn for any ε > 0; our Proposton.3 shows that δ(n) (n+ n/ )/. Probably δ(n) = n/+c n for some constant c. It seems that Theorem 4. for the non-extremal case remans vald when n s odd; the dffculty agan s on the extremal cases. Can one fnd a proof of Theorem. (n fact, only Theorem 4.) wthout usng the Regularty Lemma? Recently Pósa s conjecture has been (re)proved [5,] wthout the Regularty Lemma (thus t holds for all n n 0 wth some modest n 0 ). However, t s not clear f smlar approaches work on Theorem. or Theorem.4. Acknowledgments The authors thank the referees for ther valuable comments that mproved the presentaton. Sato s research s partally supported by the Japan Socety for the Promoton of Scence, Grant-n-Ad for Scentfc Research (C), , 004, and The Research Grant of Nhon Unversty, College of Humantes and Scences. Zhao s research s partally supported by the NSA H and H References [] S. Abbas, Spannng subgraphs of dense graphs and a combnatoral problem on strngs, Ph.D. Thess, Rutgers Unversty, 998. [] M. Agner, S. Brandt, Embeddng arbtrary graphs of maxmum degree two, J. Lond. Math. Soc. () 48 () (993) MR MR389 (94g:05063). [3] N. Alon, E. Fscher, -factors n dense graphs, Dscrete Math. 5 ( 3) (996) 3 3. MR MR38868 (97d:056). [4] J. Böttcher, M. Schacht, A. Taraz, Proof of the bandwdth conjecture of Bollobás and Komlós, Math. Ann. 343 () (009) MR MR [5] P. Châu, L. DeBaso, H. Kerstead, Pósa s conjecture for graphs of order at least 0 8, Random Structures Algorthms 39 (4) (0) [6] A. Czygrnow, H.A. Kerstead, -factors n dense bpartte graphs, Dscrete Math. 57 ( 3) (00) Kletman and Combnatorcs: A Celebraton, Cambrdge, MA, 999. MR MR (003h:0550). [7] R. Destel, Graph Theory, thrd ed., n: Graduate Texts n Mathematcs, vol. 73, Sprnger-Verlag, Berln, 005, MR MR5959 (006e:0500). [8] G.A. Drac, Some theorems on abstract graphs, Proc. Lond. Math. Soc. (3) (95) MR MR (3,856e). [9] M.H. El-Zahar, On crcuts n graphs, Dscrete Math. 50 ( 3) (984) MR MR7537 (85g:0509). [0] P. Erdős, Problem 9, n: M. Felder (Ed.), Theory of Graphs and ts Applcatons, n: Acad. Sc., Publ. House Czech, 964 (n Prague). [] P. Erdős, A. Rény, V.T. Sós, On a problem of graph theory, Studa Sc. Math. Hungar. (966) MR MR036 (36 #630). [] G. Fan, H.A. Kerstead, The square of paths and cycles, J. Combn. Theory Ser. B 63 () (995) MR MR (95k:05). [3] R.J. Gould, Advances on the Hamltonan problem a survey, Graphs Combn. 9 () (003) 7 5. MR MR (004a:0509). [4] J. Komlós, The blow-up lemma, Combn. Probab. Comput. 8 ( ) (999) 6 76, Recent trends n combnatorcs (Mátraháza, 995) MR MR68467 (000b:0508). [5] J. Komlós, G.N. Sárközy, E. Szemeréd, On the square of a Hamltonan cycle n dense graphs, n: Proceedngs of the Seventh Internatonal Conference on Random Structures and Algorthms, Atlanta, GA, 995, vol. 9, 996, pp. 93. MR MR6764 (99f:05073). [6] J. Komlós, G.N. Sárközy, E. Szemeréd, Blow-up lemma, Combnatorca 7 () (997) MR MR (99b:05083). [7] J. Komlós, G.N. Sárközy, E. Szemeréd, On the Pósa Seymour conjecture, J. Graph Theory 9 (3) (998) MR MR (99g:055). [8] J. Komlós, G.N. Sárközy, E. Szemeréd, Proof of the Seymour conjecture for large graphs, Ann. Comb. () (998) MR MR6899 (000h:0544). [9] J. Komlós, M. Smonovts, Szemeréd s regularty lemma and ts applcatons n graph theory, n: Combnatorcs, Paul Erdős s Eghty, Vol., Keszthely, 993, n: Bolya Soc. Math. Stud., vol., János Bolya Math. Soc., Budapest, 996, pp MR MR (97d:057). [0] T. Kövar, V.T. Sós, P. Turán, On a problem of K. Zarankewcz, Colloq. Math. 3 (954) MR MR (6,456a).

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